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Math Help - optimization

  1. #1
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    optimization

    Find an equation of the line through the point (2
    , 1) that cuts off the least area

    from the first quadrant.
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  2. #2
    GAMMA Mathematics
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    The line must satisfy f(2) = 1, so the slope and y-intercept will vary. We want the minimum area.

    y=mx+b with 1=2m+b,
    b=1-2m and y=mx+1-2m

    Therefore...
    y=m(x-2) + 1

    We want to minimize the area of the 1st quadrant or the area of y-intercept * x-intercept

    The x-intercept is when y=0
    0=m(x-2)+1
    x-int=\frac{-1}{m}+2

    The y-intercept is just b
    b = 1-2m

    Area must be equal to: 0.5(\frac{-1}{m}+2)(1-2m) = 0.5(\frac{-1}{m} + 2 + 2 - 4m)

    Find the derivative and set equal to zero...
    0 = \frac{0.5}{m^2} - 2
    2m^2 = 0.5
    m^2 = 0.25
    m = -0.5, .5

    Slope must equal -0.5 since a positive slope will give an infinite area in the 1st quadrant...

    AREA: 0.5(\frac{-1}{m}+2)(1-2m) = 0.5(\frac{-1}{-0.5} + 2 + 2 - 4(-0.5))
    0.5(\frac{-1}{-0.5} + 2 + 2 - 4(-0.5)) = 4
    Last edited by colby2152; December 4th 2007 at 04:54 AM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by colby2152 View Post
    Any more details to this question? Theoretically, you could use the following function that will have an area of zero...

    f(x) = 0, x \neq 2
    f(x) = 1, x = 2
    But that isn't a line...

    -Dan
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  4. #4
    GAMMA Mathematics
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    Quote Originally Posted by topsquark View Post
    But that isn't a line...

    -Dan
    Yeah, I re-read the OP's question, and stumbled to no answer at all. I'll look at it tomorrow when I am more awake.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by colby2152 View Post
    The line must satisfy f(2) = 1, so the slope and y-intercept will vary. We want the minimum area OR integral. Considering you do not need need to know calculus for this type of problem, we will avoid integrals for now...

    Back to the problem... y=mx+b with 1=2m+b,
    so b=1-2m and y=mx+1-2m

    Therefore...
    y=m(x-2) + 1

    We want to minimize the area of the 1st quadrant or the area of y-intercept * x-intercept

    The x-intercept is when y=0
    0=m(x-2)+1
    x-int=\frac{-1}{m}+2

    The y-intercept is just m
    Good up to this point. The y-intercept is 1 - 2m as you showed above.

    So the area will be
    A = \frac{1}{2} \left ( -\frac{1}{m} + 2 \right )(1 - 2m)

    By the way, it changes nothing in the answer, but you should at least deal with what happens in the case of a vertical line or horizontal line, just to be complete.

    -Dan
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  6. #6
    GAMMA Mathematics
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    Quote Originally Posted by topsquark View Post
    Good up to this point. The y-intercept is 1 - 2m as you showed above.

    So the area will be
    A = \frac{1}{2} \left ( -\frac{1}{m} + 2 \right )(1 - 2m)

    By the way, it changes nothing in the answer, but you should at least deal with what happens in the case of a vertical line or horizontal line, just to be complete.

    -Dan
    I was tired... I meant to say that the y-intercept is just b! Thanks for clearing up my work.
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  7. #7
    GAMMA Mathematics
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    Okay, I edited my original reply with a full solution. I hope the OP knows Calculus.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by colby2152 View Post
    Okay, I edited my original reply with a full solution. I hope the OP knows Calculus.
    Seems to me to be the only way to do it, so I wouldn't worry.

    -Dan
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